1.8 Solving simple quadratic equations Using spreadsheets
2. Linear relationships
2.1 Visualising linear models
2.2 Understanding gradient
2.3 The equation y=mx+c
2.4 Direct variation
2.5 Linear modelling Using spreadsheets Cumulative review (Algebra)
3. Earning money
3.1 Earning an income
3.2 Allowances, bonuses and loadings
3.3 Commission, piece work and royalties
3.4 Government allowances and pensions
3.5 Deductions and net income
3.6 Income tax Using spreadsheets
4.
4.1
Managing money
Financial calculations
4.2 Payment options
4.3 Purchasing a car
4.4 Ongoing costs of a vehicle Using spreadsheets 4.5 4.6 Fuel consumption Budgeting Using spreadsheets Cumulative review (Financial mathematics)
5. Practicalities of measurement
5.1 Significant figures
5.2 Scientific notation
5.3 Metric units of measurement
5.4 Unit prefixes in the metric system
6. Perimeter, area and Pythagoras’ theorem
6.1 Perimeter
6.2 Pythagoras’ theorem
6.3 Area of simple shapes
6.4 Area of composite figures
6.5 Perimeter and area of irregular shapes
7. Surface area, volume and capacity
7.1 Surface area of prisms
7.2 Surface area of cylinders and spheres
7.3 Volume of prisms, cylinders and spheres
7.4 Volume of pyramids and cones
7.5 Capacity
8. Time and location
8.1 Latitude and longitude
8.2 Position on Earth’s surface
8.3 Time units and intervals
8.4 World time zones
8.5 Australian time zones
9. Networks, paths and trees
9.1 Introduction to networks
9.2 Walks, paths and cycles
9.3 Trees
9.4 Minimum spanning trees
9.5 The shortest path
Cumulative review (Measurement & Networks)
10. Collecting and classifying data
10.1 Statistical investigations
10.2 Classifying data
10.3 Census or sample and self-selected sampling
10.4 Random sampling
10.5 Stratified sampling
10.6 Systematic sampling
10.7 Suitability of sample types
11. Displaying and interpreting data
11.1
11.2
11.3
11.4
11.5
11.6
11.7
11.8
11.9
Dot plots and stem-and-leaf plots
Tables and column graphs
Line graphs
Histograms and polygons
Sector graphs and divided bar charts
Cumulative frequency
Choosing a data display
Interpreting data displays
Describing the shape of datasets Using spreadsheets
12. Data analysis
12.1 Mean and mode
12.2 Median
12.3 Relative merits of mean, mode and median
12.4 Range and interquartile range
12.5 Box plots
12.6
12.7
12.8
Interpreting box plots
Standard deviation
Clusters and outliers Using spreadsheets
Cumulative review (Statistics) Exam revision
Formulas and equations Module 1 DRAFT
Lessons
• Lesson 1.1 Substitution into expressions and formulas
• Lesson 1.2 Solving linear equations
• Lesson 1.3 Solving further linear equations
• Lesson 1.4 Variables and formulas
• Lesson 1.5 Travel calculations
• Lesson 1.6 Blood alcohol content calculations
• Lesson 1.7 Medication calculations
• Lesson 1.8 Solving simple quadratic equations
Prerequisite skills
Take the diagnostic pre-test to assess your knowledge of the prerequisite skills listed below.
Diagnostic pre-test: Formulas and equations
After completing the diagnostic pre-test, brush up on your knowledge of the prerequisite skills by using the downloadable support sheets.
Support sheet: Understanding order of operations
Support sheet: Finding powers and square roots
Support sheet: Equivalent fractions
Support sheet: Understanding rates
Support sheet: Expressions and equations
Support sheet: Writing formulas
Syllabus links
• Formulas and equations (MST-11-01)
◦ A student selects and applies algebraic techniques to solve problems involving equations and formulas.
• Working mathematically (MAO-WM-01)
◦ A student develops understanding and fluency in mathematics through exploring and connecting mathematical concepts, choosing and applying mathematical techniques to solve problems, and communicating their thinking and reasoning coherently and clearly.
Lesson 1.1 Substitution into expressions and formulas
Learning intentions
By the end of this lesson, you will be able to ...
→ substitute numbers into expressions, equations and formulas for pronumerals and evaluate the result.
Substitution
• When a pronumeral such as x appears in an expression like 3x − 5, we may substitute in a number, for example 2, by replacing every instance of the pronumeral x with a 2. When substituting a number into an expression, we usually place the substituted value within brackets.
→ A multiplication symbol (×) can be included when substituting values to make the implied multiplication clear.
• After substituting in x = 2, we can evaluate the above expression as follows: 3 × (2) − 5 = 6 − 5 = 1.
In this example, we say that 2 was substituted in for x, and that the resulting value of the expression is 1.
• A mathematical formula may appear in a variety of applications. Usually, a formula has a single variable on the left-hand side of an equals sign, followed by the rest of the formula on the right-hand side. The single variable on the left-hand side is called the subject of the formula and its value is found by substituting other values into the right-hand side of the formula and evaluating.
→ The words ‘formulas’ and ‘formulae’ both mean more than one formula.
Worked example 1.1A Evaluating expressions by substituting given values for pronumerals
For a = 5, b = −2 and c = 3, find the value of:
2
Part Think Write
a Substitute 5 for a and 3 for c.
Evaluate the resulting expression using the order of operations.
b Substitute 5 for a and −2 for b.
Evaluate under the square root sign first, then find the square root.
DRAFT
c Substitute 3 for c and −2 for b When −2 is squared the answer is +4.
Evaluate the resulting expression using the order of operations.
Worked example video: Evaluating expressions by substituting given values for pronumerals
Exercise 1.1A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When asked to evaluate (or find the value of) an expression, substitute all known values into the expression.
• When substituting, put the number in brackets wherever the variable appears in the expression and then use the correct order of operations to determine the value.
• Be careful when using a calculator to ensure the correct order of operations is followed.
1 For x = 7 and y = −3, complete the following to find the value of:
a 3x + 2y = 3(7) +2(__) = __ b x − 2y = (__) − 2(−3) = __
2 For a = 2 and b = 5, find the value of: a
3 For p = 2 and q = −5, find the value of:
1 3 (p + q)
i −5(3p − 2q) j 4 − (p − q)
4 For m = 3 and n = 7, find the value of:
Worked
example 1.1B Substituting into a formula to find a value
Given that v = u + at, find v when u = 50, a = 10 and t = 6.
Think Write
Substitute 50 for u, 10 for a and 6 for t in the equation. Evaluate using order of operations or your calculator. v = u + at = 50 + 10 × 6 = 50 + 60 = 110
DRAFT
Worked example video: Substituting into a formula to find a value
Exercise 1.1B Understanding, fluency and communicating
Answers: pXXX
5 Complete the following to find t using the formula t = a + (n − 1)d when a = −3, n = 20 and d = 2 t = −3 + (20 − 1) × (__) = __
6 For each formula, find the value of the unknown variable.
a Given that V = IR, find V when I = 10 and R = 250.
b Given that V = IR, find V when I = 50 and R = 20
c Given that W = Fs, find W when F = 100 and s = 12.
d Given that s = d t , find s when d = 50 and t = 10
7 For each formula, find the value of the unknown variable.
a Given that v = u + at, find v when u = 5, a = 10 and t = 7.
b Given that s = ut + 1 2 at 2 , find s when u = −3, a = 10 and t = 5
c Given that d = 180 − 360 n , find d when n = 12
d Given that a = 2Rn n + 1 , find a when R = 8 and n = 24
8 The volume, V, of a sphere is found using the formula V = 4 3 πr 3, where r is the radius. Find, to one decimal place, the volume of a sphere with radius:
a 6 cm
b 8.5 cm
c 1.3 m
9 The amount of money in an investment that pays compound interest is given by A = P(1 + r) n. Find A when:
a P = 1000, r = 0.1 and n = 5
c P = 2340, r = 0.035 and n = 10
b P = 600, r = 0.05 and n = 20
d P = 625, r = 0.112 and n = 8
10 Pythagoras’ theorem can be written as c = √a 2 + b 2 . Find c when:
a a = 3, b = 4 b a = 5, b = 12
c a = 15, b = 20 d a = 16, b = 12
Exercise 1.1C Problem solving and reasoning
Answers: pXXX
11 Using a = 2 and b = −5, and showing all working, check whether these pairs of expressions are the same. For any pairs that are not the same, rewrite the second expression to make it the same as the first.
a 4(a + b) and 4a + 4b
c 2(3a + b) and 6a + 3b
b 4(a + b) and 4a + b
d 3a − 5(a + b) and 2a − 5b
e 3a − 5(a + b) and −2a − b f 3a − 5(a + b) and −2a − 5b
g (3a − 2b)(4a + 7b) and 12a 2 − 14ab
i (a + b) 3 and a 3 + b 3
12 Let a = 3 and b = −4.
h (3a − 2b)(4a + 7b) and 12a 2 − 8ab − 14
j (3a − 2b)(4a + 7b) and 12a 2 + 13ab − 14b 2
a Evaluate the following by substituting the values of a and b
i 4(a + b) ii a 2 + b − 9 iii 4a + 4b iv −10 + 2a
b What do you notice about the value of all these expressions?
c Does this necessarily mean that all these expressions are always equal?
d Test using other values of a and b to find which, if any, of the expressions are always equal.
13 The body mass index, b, is a measure used to determine if a person’s mass is within a recommended range. The range is 21 < b < 25
The body mass index formula is given by b = m h 2 where m = mass in kilograms and h = height in metres.
a Calculate the value of b for a person of mass 65 kg and height 1.65 m. Is the value within the recommended range?
b Peter has a mass of 100 kg and is 1.95 m tall. He states that his mass is within the recommended range. By calculating the value of b for Peter, determine if he is correct.
c Kristy has a mass of 50 kg , is 1.5 m tall and feels that she is below the recommended body mass. Is she correct? Explain using calculations.
14 To hire beach chairs a $50 deposit must be paid as well as $10 per chair hired.
a Write an algebraic expression for the total hire cost, using n to represent the number of chairs hired.
b How much would it cost to hire 6 chairs?
15 In basketball, players score points by throwing the ball through the basket, in an act known as ‘shooting’. Different points are awarded to players based on where they are standing when they shoot.
3 points Awarded to players successfully shooting from behind the three-point line
2 points Awarded to players successfully shooting from anywhere inside the three-point line
1 point Awarded to players successfully shooting when playing a foul shot
a Using appropriate variables, write an algebraic expression to represent the total number of points scored during a basketball game. Remember to carefully define your variables.
b Calculate the total number of points scored by a team during part of a practice game in which the players successfully scored 5 goals from behind the three-point line, 12 goals from inside the three-point line and 4 goals when playing foul shots.
c Find three different ways of achieving a score of 68 points.
16 Emma walks along a beach. She starts 5 kilometres from the lighthouse and travels at 6 kilometres per hour.
a Write an algebraic expression to represent Emma’s distance from the lighthouse, using t to represent the number of hours she has been walking.
b How far must Emma have walked if she has been walking for half an hour?
c For how long must Emma have walked if she is 14 km from the lighthouse?
Exercise 1.1D Challenge
Answers: pXXX
17 Substitute a = 3, b = 4 and C = 56 into the following expression. Write your answer correct to one decimal place.
√a 2 + b 2 − 2ab cos (C)
Lesson 1.2 Solving linear equations
Learning intentions
By the end of this lesson, you will be able to ...
→ solve linear equations.
Solving linear equations
• An algebraic equation consists of two algebraic expressions that are equal to each other. The aim of solving an equation is to find the value or values of the pronumeral that make the equation a true statement.
• To solve a linear equation:
1. Keep the equation balanced by performing the same operation to both sides of the equation; that is:
• add the same number or term to both sides of the equation
• subtract the same number or term from both sides of the equation
• multiply both sides of the equation by the same number or term
• divide both sides of the equation by the same number or term.
2. Continue until you obtain the simplest equivalent equation with the pronumeral on one side of the equation and a number on the other. For example, to solve x + 4 = 10, we subtract 4 from both sides of the equation to obtain the solution x = 6 x + 4 = 10 (− 4) x = 6
Worked example 1.2A Determining whether a given value is a solution to an equation
Substitute the given value into the equation to see if it is a solution to the equation.
a 3x − 9 = 36 [x = 11] b 49 − 5x = 29 [x = 4]
Part Think
a Substitute x = 11 into 3x − 9 and simplify to get 24
But 24 ≠ 36, so x = 11 is not a solution.
DRAFT
b Substitute x = 4 into 49 − 5x to get 29
This means x = 4 is a solution.
Write
3x − 9 = (3 × 11) − 9 = 33 − 9 = 24 ≠ 36
x = 11 is not a solution.
49 − 5x = 49 − (5 × 4) = 49 − 20 = 29
x = 4 is a solution.
Worked example video: Determining whether a given value is a solution to an equation
Exercise 1.2A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• To determine whether a given value is a solution to an equation, substitute the value into the expression.
→ If both sides of the equation are the same, then that value is a solution.
→ If the two sides are not the same, then it is not a solution.
1 Complete the following to determine whether:
a x = 5 is a solution to 7x − 3 = 28.
7x − 3 = (7 × __) − 3 = 35 − 3 = __
x = 5 is
b a = 8 is a solution to 7 + 3a = 31
7 + 3a = 7 + (3 × __) = 7 + __ = 31 a = 8 is
2 Substitute the given value for the pronumeral into the equation to determine if it is a solution to the equation.
a 4x + 8 = 60 [x = 13] b 5x − 9 = 11 [x = 3] c 3x + 4 = 13 [x = 3]
d 9 − 2p = −15 [p = 12] e 10 + 5m = −20 [m = −6] f 22 + 3c = 46 [c = 7] g 6d − 5 = 31 [d = 9] h 11y − 72 = −193 [y = −11] i 14q + 18 = 116 [q = 7]
Worked example 1.2B Solving two-step linear equations
Solve these equations.
a 5x + 7 = 37 b 8x − 9 = 79
Part Think Write
a Subtract 7 from both sides of the equation and simplify the result.
Divide both sides of the equation by 5 and simplify the result.
Check your answer by substituting x = 6 back into the original equation.
b Add 9 to both sides of the equation and simplify the result.
Divide both sides of the equation by 8 and simplify the result.
Check your answer by substituting x = 11 back into the original equation.
Worked example video: Solving two-step linear equations
Exercise 1.2B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Perform the same operation on both sides of the equation to keep the equations equivalent from step to step.
• Always substitute the answer back into the equation to check it is correct. Answers are not always whole numbers.
3 Fill in the blanks to solve these equations.
a 7x − 3 = 32
7x − 3 + __ = 32 + 3
4 Solve the following equations.
a 3x − 9 = 3
b 4p + 8 = 24
c 6x − 2 = 82 d 6x + 7 = 61
e 4p + 12 = 52 f 5y − 6 = 44
g 2x − 32 = 16 h 11p − 10 = 34
i 7m + 8 = 50 j 4q − 21 = 115
k 8s − 12 = 252 l 10r − 30 = 100
m 7t − 8 = 27
o 4m + 9 = 73
11s + 16 = 115
Worked example 1.2C Solving linear equations containing a negative pronumeral term
Solve the following linear equations.
a 9 − 7x = 5 b 17 = 8 − 4x
Part Think
a Subtract 9 from both sides and simplify the result. Divide both sides by −7 and simplify to get x = 4 7
Check your answer by substituting x = 4 7 back into the original equation.
Write
Part Think Write
b Subtract 8 from both sides of the equation and simplify the result. Divide both sides of the equation by −4 and simplify. The answer is x = − 9 4 . The result can also be written as a mixed number as x = −2 1 4 . Check your answer by substituting x = − 9 4 back into the original equation.
Worked example video: Solving linear equations containing a negative pronumeral term
Exercise 1.2C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When solving a linear equation, the goal is to have the pronumeral on one side of the equation and the numbers on the other side. Perform the same operation on both sides of the equation to isolate the variable.
• The solving process is the same whether the equation appears as 9 − 7x =
x + 9 = 5: we start by subtracting 9 from each side. Similarly, 17 = 8 − 4x can be solved in the same way as 8 − 4x = 17 or −4x + 8 = 17: we subtract 8 from each side.
Add 5 to both sides of the equation and simplify the result.
Multiply both sides of the equation by 3 and simplify the result to get m = 9.
Check your answer by substituting m = 9 back into the original equation.
Worked example video: Solving a linear equation
Exercise 1.2D Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Isolate the variable on one side of the equation by applying operations to both sides of the equation. Often this means that you can continue to use the same technique of adding or subtracting first before multiplying or dividing.
7 Fill in the blanks to solve the equation.
x 5 + 3 = 7
x 5 + 3 − __ = 7 − __ x 5 = __ x 5 × __ = 4 × 5 x =
8 Solve each of these equations for x a x 2 + 3 = 8
x 6 + 3 = −4
x 2 − 1 = 4
x 7 − 2 = 4
9 Check the given solution by substitution and state whether or not it is correct.
a 2x + 8 = 15 [x = 7]
b 7 − 5x = 9 [x = − 2 5 ]
c −15 = 6 − 7x [x = 3] d x 5 − 3 = 6 [x = 9 5 ]
x 5 + 2 = −3
Worked example 1.2E Solving a linear equation resulting from substitution
Given y = 5x − 3, find x when y = −18
Think Write
Substitute the value for y, giving an equation that can be solved. Add 3 to both sides of the equation and simplify the result.
Divide both sides of the equation by 5 and simplify the result to get x = −3
Check your answer by substituting x = −3 back into the original equation. y = 5(−3)−3 = −15 − 3 = −18
Worked example video: Solving a linear equation resulting from substitution
Exercise 1.2E Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Some equations have more than one unknown. Substitute any variables that you know the value of and then solve for the value of the desired variable in the usual way.
• Make sure you substitute a given value for the correct variable!
10 Given y = 7x + 11, fill in the blanks in the following to find x when y = −3. y = 7x + 11 = 7x + 11
−3 − 11 = 7x + 11 − __ = 7x 14 7 = 7x □ x = ___
11 Substitute the value for y to solve for x
a Given y = 3x − 5, find x when y = −14
b Given y = 4x + 2, find x when y = 11
c Given y = 7 − 5x, find x when y = 8
d Given y = 4 − 3x, find x when y = −3.
DRAFT
e Given y = 5 − 7x, find x when y = −5
f Given y = 3x − 5, find x when y = 0.
Worked example 1.2F Developing and
solving an
equation from a worded description
When five is subtracted from twice a number, the answer is three. Write an equation and solve it to find the original number.
Think Write
Write an equation based on the description.
Let x represent the original number.
‘Twice a number’ means to multiply x by 2 and ‘five is subtracted’ means to subtract 5.
‘The answer is three’ means to write the expression equal to 3 Let the original number be x.
Solve the equation.
Add 5 to both sides of the equation and simplify the result.
Divide both sides of the equation by 2 and simplify the result to get x = 4
Check your answer by substituting x = 4 back into the original equation. 2(4)−5 = 8 − 5 = 3
Worked example video: Developing and solving an equation from a worded description
Exercise 1.2F Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Begin by defining the variable so that you can ensure the answer makes sense.
• Form the equation using the information in the question. Pay attention to the order of operations applied to the unknown number.
• After solving, make sure the answer makes sense in the context of the question.
12 By first forming an equation, solve each of the following worded questions.
a When five is subtracted from three times a number, the answer is thirteen. Find the original number.
b When seven is subtracted from four times a number, the answer is thirteen. Find the original number.
c When nine is added to five times a number, the answer is thirty-four. Find the original number.
DRAFT
d When eight is added to seven times a number, the answer is twenty-nine. Find the original number.
Exercise 1.2G Problem solving and reasoning
Answers: pXXX
13 The sum of two consecutive integers is 275
a If the first number is x, write an algebraic expression for the second number.
b Write an equation in terms of x to represent this problem.
c Solve the equation and find the two numbers.
14 For each of these problems, first form an equation then solve.
a Two consecutive integers add to 127. Find the number s.
b Three consecutive integers have a sum of 27. Find the smallest of the three numbers.
c Four consecutive integers have a sum of −6. Find the largest of them.
d Three consecutive even integers have a sum of 132. Find the smallest of the three numbers.
15 Tyler is saving to buy a cricket bat that costs $235. He is able to save $28 per week and currently has $67. Follow the steps below to work out when he will be able to buy the bat.
a Choose a pronumeral to represent the unknown quantity in the problem.
b Use the pronumeral to write an equation to represent the problem.
c Solve the equation and write your answer.
16 For each of these problems, form an equation to solve. Remember to clearly define the pronumeral you use to represent the unknown variable.
a Rachel bought three equally priced dresses online for a total cost of $196, which included the delivery charge of $19. What is the cost of each dress?
b The cost of hiring a tennis court is $9 plus $14 per hour. For how many hours can you hire the tennis court if you and your friends have a total of $85?
c Andrew and Todd scored a total of 37 goals between them in a basketball match. Todd scored five more goals than Andrew. How many goals did Andrew score?
d Tina has a budget of $1200 for her birthday party. The cost of hiring a local hall is $325 and catering is $28 per person. What is the maximum number of people that can attend?
17 The length of a rectangular playing field is 14 m lon ger than its width.
a If w represents the width of the field, write an expression for the length of the field in terms of w
b Form an equation for the perimeter of the field in terms of w
c If the perimeter of the field is 156 m, solve an equation to find the dimensions of the playing field.
18 In an AFL match, players can score points for their team in two ways. They can kick a goal, which is worth six points, and they can score a behind, which is worth one point.
a Using appropriate variables, write an algebraic equation to calculate the total number of points scored during an AFL game. Remember to carefully define your variables.
b Calculate the number of points scored if a team kicks 12 goals and 3 behinds.
c Calculate the number of points scored by both teams in each of the following games and identify the winner. Show all working.
i Kangaroos: 6 goals and 12 behinds, Swans: 7 goals and 8 behinds
ii Eagles: 23 goals and 3 behinds, Suns: 17 goals and 25 behinds
d The scoring in the following games is incomplete. Calculate the number of behinds scored in each of the games. Show all working.
i Cats: total points was 97, number of goals was 15
ii Power: total points was 128, number of goals was 19
e The scoring for the games below is also incomplete. Calculate the number of goals scored in each of the games. Show all working.
i Saints: total points was 87, number of behinds was 9
ii Hawks: total points was 131, number of behinds was 17
f There has been some discussion about the possibility of changing the scoring system so that players receive three points if they hit the post instead of kicking a goal. How would this change your equation?
g If the change described in part f was implemented and a team scored 15 goals and 14 behinds, and hit the post a number of times for a total score of 119 points, how many times did the ball hit the goalpost?
Exercise 1.2H Challenge
Answers: pXXX
19 Hayley, Davina and Marcel go fishing. Davina catches four more fish than Marcel, and Hayley catches twice as many fish as Marcel.
a Form an equation to find the number of fish each person catches. Explain why this problem cannot be solved.
b Hayley believes they caught a total of 65 fish. Does this seem reasonable? Explain.
20 Equations with two unknowns can have an infinite number of solutions. The solutions will be a pair of numbers for x and y. For example, x = 3y − 7
a Verify that when x = 5, y = 4.
b Find another pair of integers for x and y which make the equation true.
c Find all the pairs of integers for x and y which make the equation tr ue, where 0 ≤ x ≤ 20.
Related resources
1.2 Quick quiz
Worksheet: Solving simple linear equations
Investigation: Household bills
Lesson 1.3 Solving further linear equations
Learning intentions
By the end of this lesson, you will be able to ...
→ solve equations after substituting numbers for pronumerals into formulas.
Solving more complex linear equations
• In the previous lesson, each equation could be solved by performing a simple addition or subtraction, followed by a simple multiplication or division. Solving the equations in this lesson will require different sequences of operations and different types of terms will appear.
• Some linear equations have one or more terms that are fractions. Adding or subtracting a number from both sides of an equation follows the same process whether that number is a whole number or a fraction.
• If the pronumeral of interest appears in the numerator of a fraction, multiply both sides of the equation by the fraction’s denominator, then expand any brackets and simplify. Once the pronumeral is no longer part of a fraction, solve the resulting equation as normal.
• Sometimes the pronumeral of interest will appear on both sides of an equation. In this situation, the first step is to remove one of the pronumeral terms from both sides of the equation by adding or subtracting. Once the pronumeral only appears on one side of the equation, solve the resulting equation as normal.
• After finding a solution to an equation, the result should always be checked by substituting it back in. If both sides of the evaluated equation are not equal, then the result was incorrect and should be reviewed.
Worked example 1.3A Solving linear equations involving fractions
Solve the following for x a
Part Think Write
a Add 1 5 to both sides, then simplify.
b Subtract 2 3 from both sides. Divide both sides by 2, then simplify.
Worked example video: Solving linear equations involving fractions
Exercise 1.3A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• A fraction can be added or subtracted from both sides of an equation, just like an integer can.
1 Complete the following to solve for x x − 2 5 = 7 5
x − 2 5 + 2 5 = 7 5 + □ □ x = □ □ or x = 1 □ 5
2 Solve the following linear equations for x
Worked example 1.3B Solving more linear equations involving fractions
Solve for x
a x 3 = 2 5 b 4x + 3 5 = −2
Part Think Write
a Multiply both sides by 3, then simplify.
b 1 Multiply both sides of the equation by the denominator, 5
2 Subtract 3 from both sides of the equation, then divide both sides by 4.
3 Simplify the solution (if required).
Worked example video: Solving more linear equations involving fractions
Exercise 1.3B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• An expression written as the fraction 1 n multiplied by a collection of terms enclosed by brackets can be rewritten so that the terms inside the brackets appear as the numerator of the fraction. For example, 1 4 (2x − 7) = 2x − 7 4 .
3 Complete the following to solve for x
x 5 = 7 8 x 5 × 5 = 7 8 × __ x = □ □ or x = __ 3 8
4 Solve the following for x a x 3 =
5 Complete the following to solve for x. Write your answer as a mixed fraction.
Solving linear equations with a pronumeral term on both sides
Solve these equations.
a 5x + 2 = 3x − 5
Part Think Write
a First, subtract 3x from both sides of the equation. Then, subtract 2 from both sides. Finally, divide both sides by 2, then simplify.
Part Think Write
b First, add 2x to both sides of the equation. Then, subtract 11 from both sides.
Finally, divide both sides by 3, then simplify.
Worked example video: Solving linear equations with a pronumeral term on both sides
Exercise 1.3C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When solving equations with pronumerals on both sides, it may be necessary to not only add and subtract numbers from both sides, but also to add and subtract pronumerals from both sides.
• It does not matter which side of the equation you remove a pronumeral term from but sometimes one choice is more convenient than the other. For example, if an equation has 3x on the left-hand side and 2x on the right-hand side, you may choose to subtract 2x from both sides so that the surviving pronumeral term is positive.
8 Solve the following equations with integer solutions.
9 Solve these equations, giving your answer as a mixed fraction where appropriate.
DRAFT
j 5a + 3 = a − 3
− 7 = 5a + 15
Worked example 1.3D Checking solutions to linear equations
Substitute the value given in square brackets to check whether it is the solution to each equation.
a 2x − 5 = 10 − 3x [x = 3]
b 5x + 2 = 2x − 7 [x = 2]
Part Think Write
a Substitute x = 3 into the left-hand side of the equation and evaluate, then do the same for the right-hand side.
Both sides of the equation are equal to 1, so x = 3 is a valid solution.
b Substitute x = 2 into the left-hand side of the equation and evaluate, then do the same for the right-hand side.
The left-hand side is 12 and t he right-hand side is −3, so x = 2 is not a valid solution.
Worked example video: Checking solutions to linear equations
2x − 5 = 2(3) − 5 = 6 − 5 = 1
10 − 3x = 10 − 3(3) = 10 − 9 = 1 x = 3 is a solution.
5x + 2 = 5(2) + 2 = 10 + 2 = 12
2x − 7 = 2(2) − 7 = 4 − 7 = −3 x = 2 is not a solution.
Exercise 1.3D Understanding, fluency and communicating
Helpful hints
• When checking a solution, substitute the value into the equation and check that the two sides of the equation are equal. If they are not equal, the given value is not a valid solution.
10 Complete to show that x = 3 is a solution of 7 − 3x = x − 5 LHS: 7 − 3x = 7 − 3(__) = 7 − __ = __ RHS: x − 5 = __ − 5 = __
Hence, x = 3 __ a valid solution.
DRAFT
11 Substitute the value given in square brackets to check whether it is a solution to the equation each time. a 3x + 9 = 4 − 2x [x = −1]
Worked example 1.3E Solving a problem by first forming an equation to solve
If one more than twice a number is five more than the number, what is the number?
Think Write
2x + 1 is one more than twice the number.
x + 5 is five more than the number.
Form the equation using 2x + 1 and x + 5
Let the number be x
The equation becomes:
The number is 4.
Worked example video: Solving a problem by first forming an equation to solve
Exercise 1.3E Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When converting a worded problem into an equation, be sure to carefully define any pronumerals used.
12 Form an equation and then solve it to find the number in each of these problems.
a Three more than twice a number is seven more than the number.
b Eight more than three times a number is four more than the number.
c Six more than twice a number is four more than four times the number.
d Nine less than five times a number is three less than twice the number.
Exercise 1.3F Problem solving and reasoning
Answers:
13 A formula for converting temperature is C = 5 9 (F − 32) where C represents the temperature in degrees Celsius ( °C) and F represents the temperature in degrees Fahrenheit ( °F).
a For C = 30, write the formula as a linear equation involving the variable F
DRAFT
b Solve your linear equation to find F
c What is the equivalent temperature in degrees Fahrenheit for a temperature of 30 °C?
d Solve a linear equation to find the equivalent temperature in degrees Fahrenheit for a temperature of 18 °C
e Which temperature is higher: 25 °C or 76 °F? Show steps of working to justify your answer.
14 The number of biscuits in a packet is unknown. However, Lisa can fill the family biscuit barrel with six more biscuits than the contents of two packets, or five less biscuits than the contents of three packets.
a If n represents the number of biscuits in a packet, which equation (A, B, C or D) best matches this situation?
A 2n − 6 = 3n − 5
B 2n − 6 = 3n + 5
C 2n + 6 = 3n − 5
D 2n + 6 = 3n + 5
b Solve the equation to find the number of biscuits in a packet.
15 Kristen and Daniel have the same amount of money. Kristen buys 5 kg of cherries and has $2 left over. Daniel buys 2 kg of cherries and has $20 left over.
a Write an equation to represent this situation.
b Solve the equation to find the cost of 1 kg of cherries
16 James and Sue have the same amount of money. James buys seven sushi rolls and has $1.50 left over. Sue buys four sushi rolls and has left over. Form an equation and then solve it to find the cost of a sushi roll.
17 The average of Hayden’s four test results for the semester is 78%. He remembers that three of his test results were 91%, 69% and 88%. Form a n equation and then solve it to work out the result he received in the fourth test.
18 For three consecutive even integers, the sum of the two smaller numbers is equal to six more than the largest number. Form an equation and then solve it to find the integers.
Exercise 1.3G Challenge
Answers: pXXX
19 Solve the following linear equations for x.
20 Solve the following problems by first forming an equation.
a A man is currently three times as old as his son. In 11 year s from now, he will be twice as old as his son will be then. How old is his son now?
b At present Jen is 8 year s older than Wesley. In 3 year s from now, Jen’s age will be double Wesley’s age. How old is Wesley?
c The sum of Kelly and Sam’s ages is 45 year s. If Kelly’s age was doubled, it would be 5 year s more than three times Sam’s age. How old is Sam?
21 The formula A = 2πrh + 2πr 2 will be used in a later module. Solve for h when A = 300 and r = 5.5. Write your answer correct to two decimal places.
Related resources
1.3 Quick quiz
Investigation: Costs at a cinema
Lesson 1.4
Variables and formulas
Learning intentions
By the end of this lesson, you will be able to ...
→ change the subject of a formula that is a linear equation.
Worksheet: Solving equations with a pronumeral term on both sides
Finding the value of a variable in a formula
• A formula is an equation that relates two or more variables according to a rule. Each variable can be represented by a pronumeral.
• If the subject of a formula is a pronumeral, its value can be found by substituting numbers in for any remaining pronumerals, as in Lesson 1.1 Substitution into expressions and formulas. To find the value of a pronumeral that is not the subject of the formula, there are two approaches.
→ Substitute the known values into the formula and solve the resulting equation.
→ Rearrange the formula, so that the subject of the formula is the pronumeral to be found and then substitute any values that are known. Rearranging a formula to change its subject follows the same rules as those used for solving equations.
Worked example 1.4A Substituting
known values into a formula and solving the resulting equation
a Find the value of t when v = 117, u = 5, a = 8 and v = u + at
b Find the value of N when R = 23, I = 4 and R = I N + 1.
Part Think Write
a Substitute v = 117, u = 5 and a = 8 into the formula.
Solve the resulting equation by subtracting 5 from both sides then dividing both sides by 8
b Substitute R = 23 and I = 4 into the formula.
Solve the resulting equation by subtracting 1 from both sides, multiplying both sides by N and then dividing both sides by 22
v = u + at
117 = 5 + 8t 112 = 8t 112
8 = t t = 14
R = I N + 1
23 = 4 N + 1
22 = 4 N
22N = 4 N = 4 22 N = 2 11
Worked example video: Substituting known values into a formula and solving the resulting equation
Exercise 1.4A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Start by substituting all known values into the formula, then solve the resulting equation to make the desired pronumeral the subject.
• If the pronumeral of interest appears in the denominator of a fraction, multiply both sides of the equation by this denominator.
1 If v = u + at, find:
a t when v = 102, u = 18 and a = 7
2 Use the formula d = 1 2 ct to find the value of:
DRAFT
a c when d = 100 and t = 8
3 If M = 5K 18 , find K when M = 15
4 If V 2 = gR, find R when V = 12 and g = 10
5 If b = m h 2 , find m when b = 23 and h = 1.63
6 Use the formula I = Prn to find the value of:
a P when I = 19 500, r = 0.03 and n = 25
b a when v = 54, u = 12 and t = 14
b t when d = 320 and c = 16
b n when I = 2100, P = 5000 and r = 6 100
7 Given s = ut + 1 2 at 2, find u when s = 360, t = 8 and a = 10.
8 If S = n 2 (a + l), find a when S = 560, n = 20 and l = 53.
9 If S = n 2 (2a + (n − 1)d), find the value of a when S = 610, d = 3 and n = 20
10 Use the formula A = 180 − 360 n to find n when A = 120
11 Use the formula a = 2Rn n + 1 to find the value of:
a R when a = 12 and n = 24
12 Use the formula I = E R + r to find the value of:
a E when I = 8, R = 15 and r = 3
Worked
b R when a = 19.5 and n = 39
b R when I = 2, E = 24 and r = 3
example 1.4B Changing the subject of a formula containing more than one pronumeral
Make x the subject of each formula below. a x − 5 = r
Part Think Write
a Add 5 to both sides of the equation.
b Divide both sides of the equation by m and simplify.
c Divide both sides of the equation by z and simplify.
Equivalently, multiply both sides of the equation by 1 z .
Worked example video: Changing the subject of a formula containing more than one pronumeral
Exercise 1.4B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Apply the same rules for solving equations to rearrange the formula so that the desired pronumeral is the subject. Any other pronumerals can be treated just like numbers!
DRAFT
13 Make x the subject of each of these formulas.
a x + 2 = y b x + 5 = y c y = x + 8 d 5x = p e 7x = q f r = 2x g x 3 = t h x 5 = r i m = x 8 j x − 3 = p
14 Make x the subject of each formula below. a x + y = z
=
15 Make x the subject of each formula below.
Worked example 1.4C Changing the subject of a formula (two steps)
Make x the subject of each formula below.
a y = 2x + 3 b m = 4 − 3x c y = mx + b
Part Think Write
a Subtract 3 from both sides of the equation, then divide both sides by 2.
b Subtract 4 from both sides of the equation, then divide both sides by −3. The expression −(m − 4) is the same as −1(m − 4) or (−1 × m − 1 × −4) or m + 4.
c Subtract b from both sides of the equation, then divide both sides by m
Worked example video: Changing the subject of a formula (two steps)
DRAFT
Exercise 1.4C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Apply the same rules for solving equations to rearrange the formula so that the desired pronumeral is the subject. Any other pronumerals can be treated just like numbers!
16 Make x the subject of each formula below.
y = 3x + 2
17 Make y the subject of each of these equations. a a + by = c b p + qy = r
− b = cy
xy − p = r
x − ky = t
+ by + r = 0 h k = r − py
18 Given R = I N + 1, find the value of N when:
a R = 3 and I = 7
19 Use t = √ l g to find l when t = 1.3 and g = 10.
20 Use r = 3√ V k to find V when r = 3 and k = 4.2
b R = 11 and I = 5
21 Use the formula E = 1 2 mv 2 to find the value of v when E = 135 and m = 2.8. Give your answer correct to one decimal place.
Exercise 1.4D Challenge
Answers: pXXX
22 Make r the subject of the following formulas:
a V = 4 3 πr 3
b S = V0(1 − r) n
Related resources
1.4 Quick quiz
DRAFT
Worksheet: Changing the subject of a formula
Worksheet: Solving more linear equations
Investigation: Balanced linear equations
Lesson 1.5
Travel calculations
Learning intentions
By the end of this lesson, you will be able to ...
→ solve problems related to travel, including calculations relating speed, distance and time.
Average speed
• The formula for the average speed of an object is given below. This formula can be used to calculate the average speed, distance travelled or time taken.
Average speed = distance travelled time taken
→ This is usually written as S = D T , where S is the average speed, D is the distance travelled and T is the time taken.
→ The formula can be rearranged using inverse operations to get D = S × T and T = D S
Total stopping distance
• When drivers react to situations in which they need to stop a motor vehicle, the total stopping distance of the vehicle depends on the distance travelled while the driver reacts and the distance travelled after applying the brakes.
• The reaction-time distance is the distance travelled in the time it takes a driver to react to a situation; that is, to realise there is a problem and move their foot to the brake. The usual reaction time for drivers unaffected by alcohol, drugs or fatigue has been found to be about 2.5 seconds.
• The braking distance is the distance a car travels after the brakes have been applied. This distance depends on (the square of) the speed of the car.
→ Factors such as the condition of the road (wet or dry) and the condition of the car’s brakes and tyres, as well as the state of the driver, all have an effect on the total stopping distance.
→ The braking distance is a function of the square of the speed of the car. For a car with good brakes and tyres, travelling in dry conditions on a good road, the relationship can be approximated by the formula d = 0.01v 2, where d is the braking distance in metres and v is the speed of the car in km/h
→ For the same car travelling on a slippery road, the formula for braking distance becomes d = 0.014v 2
→ The distance a car travels in the time it takes to stop is:
Total stopping distance = reaction-time distance + braking distance
Converting units
• Use conversion factors to convert between units of distance and time.
→ Multiply to convert from a larger unit to a smaller unit.
→ Divide to convert from a smaller unit to a larger unit.
Worked example 1.5A Calculating average speed, distance and time
a A car travels 232 km in 4 hours and 17 minutes. Calculate its average speed, correct to the nearest whole number.
b A train averages 83 km/h for 2 hou rs and 24 minutes. How far does it travel?
c If a motorcyclist can average 52 km/h, how long will it take her to travel 34 km, correct to the nearest minute?
Think Write
a
Convert 17 minutes to hours. Add the result to 4 hou rs to find the total time in hours.
Substitute the values for distance (D) and time (T ) into the formula S = D T .
Simplify the result and round to the nearest whole number.
b Convert 24 minutes to hours.
Substitute the values for speed (S) and time (T ) into the formula D = S × T
Simplify the result.
c Substitute the values for distance (D) and speed (S ) into the formula T = D S .
Simplify the result.
Convert the result from hours to minutes by multiplying by 60 Round to the nearest minute.
DRAFT
17 min = (17 ÷ 60) h = 0.283 33… h
4 h 17 min = 4.283 33… h
S = D T
S = 232 km 4.283 33… h = 54.16… km/h
≈ 54 km/h
24 min = (24 ÷ 60) h = 0.4 h
D = S × T
D = 83 km/h × 2.4 h = 199.2 km
T = D S T = 34 km 52 km/h = 0.6538… h
0.6538… hours = (0.6538…× 60) min = 39.23… min ≈ 39 min (to the nearest minute)
Worked example video: Calculating average speed, distance and time
Exercise 1.5A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When using the average speed formula (S = D T ), or any rearrangements of it, include the units in your working to help keep track of the units of the result.
• Read questions carefully to make sure you round results when required.
1 Calculate the average speed for these distances and times.
a 185 km is travelled in 4 hours
c 154 km is travelled in 3 hours and 15 min
2 Calculate the distance travelled in:
a 3 hours and 30 min at an average speed of 64 km/h
b 3 hours and 20 min at an average speed of 56 km/h
c 5 hours and 47 min at an average speed of 82 km/h
d 2 hours and 13 min at an average speed of 75 km/h
3 How long will it take to travel:
a 486 km at 60 km/h
c 365 km at 82 km/h
b 720 km is travelled in 9 hours and 50 min
d 272 km is travelled in 4 hours and 35 min
b 298 km at 74 km/h
d 88 km at 95 km/h?
Worked example 1.5B Converting units of speed
Convert:
a 65 km/h to m/s (to one decimal place) b 9.8 m/s to km/h.
Part Think Write
a Write 65 km/h as a fr action.
Substitute these conversion factors to convert to the desired units:
1 km = 1000 m
1 h = 60 × 60 s
Simplify the result and write with the units m/s
b Write 9.8 m/s as a fr action.
Substitute these conversion factors to convert to the desired units:
DRAFT
1 1000 km = 1 m 1 60 × 60 h = 1 s.
65 km/h = 65 km 1 h = 65 × 1000 m 60 × 60 s ≈ 18.1 m/s
Simplify the result and write with the units km/h 9.8 m/s = 9.8 m 1 s = 9.8 × ( 1 1000 km) 1 60 × 60 h
Worked example video: Converting units of speed
= ( 9.8 1000 km) ÷ ( 1 60 × 60 h)
= 9.8 × 60 × 60 km 1000 h = 35.28 km/h
Exercise 1.5B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• When converting units of speed, include the units of distance and time in your calculations. You can then substitute the converted units into the expression and simplify to find the result.
4 Complete the following to convert the units of speed.
a 70 km/h = □ km □ h = □ × □ m □ × □ s ≈ __ m/s (to one decimal place)
b 8 m/s = □ m □ s = □ × ( 1 □ km) 1 □ × □ h = ( □ □ km) ÷ ( 1 □ × □ h) = □ × □ × □ km □ h = __ km/h
5 Convert each of the following to m/s correct to one decimal place.
a 45 km/h b 76 km/h c 110 km/h
6 Convert each of the following to km/h a 15 m/s b 12.5 m/s c 25 m/s
Worked example 1.5C Calculating reaction-time distance
Calculate the reaction-time distance correct to the nearest whole number for a car travelling at 60 km/h Assume a reaction time of 2.5 s
Think Write
Convert the speed to m/s
Use the formula D = S × T where T = 2.5 s, where T is the time it takes the driver to react.
Multiply the speed in m/s by the reaction time of 2.5 s.
Worked example video: Calculating reaction-time distance
60 km/h = 60 km 1 h = 60 × 1000 m 60 × 60 s
Exercise 1.5C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Make sure the units of the reaction time and the unit of time in the denominator of the speed are the same. When you multiply the speed by the reaction time, the units of time cancel, and you are left with the unit of distance from the numerator of the speed.
7 Calculate the reaction-time distance for a car travelling at 80 km/h. Assume a reaction time of 2.5 s
8 Calculate the reaction-time distance for a car travelling at 100 km/h. Assume a reaction time of 2.5 s.
9 Consider again the results of questions 7 and 8 .
a Use these results to draw a (straight-line) graph of the relationship between reaction-time distance (m) and speed (km/h), given a reaction time of 2.5 seconds.
b Use the graph to estimate the reaction-time distances for cars travelling at these speeds.
i 120 km/h
ii 45 km/h
c If the speed of a car increases by 10 km/h, what is the increase in the stopping distance?
Worked example 1.5D Calculating braking distance
Calculate the braking distance for a car travelling in dry conditions at 60 km/h
Think Write
Substitute the value of v = 60 into the dry conditions formula, d = 0.01v 2
Worked example video: Calculating braking distance
Braking distance = 0.01v 2 = 0.01 × 60 2 = 36 m
Exercise 1.5D Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• The wording of the question will tell you which braking distance formula to use.
→ If a question uses words like ‘dry’ or ‘good’ to describe the conditions, use the formula d = 0.01v 2
→ If a question uses words like ‘wet’, ‘poor’ or ‘slippery’ to describe the conditions, use the formula d = 0.014v 2
→ Some questions will give you a new formula to use. Apply the same technique as above with any other braking distance formula.
10 Calculate the braking distance for a car travelling at 80 km/h in good conditions.
11 A car is travelling at 100 km/h
a Calculate the braking distance if the car is travelling in good conditions.
b What is the braking distance if the car is travelling in wet conditions?
c What is the difference between the braking distances at 100 km/h in good conditions and in wet conditions?
Worked example 1.5E Calculating total stopping distance
Find the total stopping distance, to the nearest metre, for a car travelling at 70 km/h in good conditions, assuming a reaction time of 2.5 s.
Think Write
Convert 70 km/h to m/s.
Apply the formula D = S × T to find the reactiontime distance, where S is the speed in m/s and T is the reaction time.
Apply the formula d = 0.01v 2 to find the braking distance, where v is the speed in km/h and d is the distance in metres.
Add the reaction-time distance to the braking distance to find the stopping distance.
Worked example video: Calculating total stopping distance
70 km/h = 70 km 1 h = 70 × 1000 m 60 × 60 s
Reaction-time distance = 70 × 1000 m 60 × 60 s × 2.5 = 48.611... m
Braking distance = 0.01 × 70 2 m = 49 m
Total stopping distance = 48.611... m + 49 m = 97.611... m ≈ 98 m
Exercise 1.5E Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Pay attention to the units of speed used in each formula. In the example above, the speed is in m/s to find the reaction-time distance but in km/h to find the braking distance.
12 Complete the following to find the total stopping distance for a car travelling at 90 km/h in good conditions, assuming a reaction time of 2.5 seconds.
90 km/h = □ km □ h = □ × □ m □ × □ s
Reaction-time distance = □ × □ m □ × □ s × □ s = □ m
DRAFT
Braking distance = 0.01 × □ 2 m = □ m
Stopping distance = □ m + □ m = __ m ≈ __ m (to the nearest metre)
13 Find the total stopping distance for a car travelling at 110 km/h in good conditions, assuming a reaction time of 2.5 seconds.
Exercise 1.5F Problem solving and reasoning
Answers: pXXX
14 When David is affected by fatigue, he has a reaction time of 3.5 seconds. When not affected by fatigue, his reaction time is only 2.5 seconds. He is driving at 60 km/h
a What is the reaction-time distance when David is affected by fatigue?
b What is the difference between the stopping distances of when David is affected by fatigue and when he is not?
15 When a given driver is affected by alcohol, they have a reaction time of 4.5 seconds. What difference will this make to the reaction-time distance of a car travelling at 100 km/h compared with the usual reaction time of 2.5 seconds?
16 The reaction-time distance, d m, for a car travelling at v km/h, assuming a reaction time of t s, can be approximated using the formula d = 0.28 vt. Use this formula to check your answers for questions 14 and 15.
17 A car is travelling in good conditions, and the driver has the usual reaction time of 2.5 seconds. If the speed of the car increases from 50 km/h to 60 km/h, what is the increase in total stopping distance?
18 Find the stopping distance for a car travelling at 70 km/h in good conditions. Assume a reaction time of 2.5 seconds and use the formula d = 0.7v + 0.01v 2, where d is the stopping distance in metres and v is the speed in km/h.
19 Another formula for the stopping distance is d = 0.7v + 0.01v 2, where d is in metres and v is the speed in km/h.
a Use this formula to complete the following table.
distance (m) 0.7 × 20 + 0.01 × 20 2 = 18
b Use the information in the completed table from part a to graph the relationship between speed and stopping distance. (Join the points with a smooth curve.)
c Using your graph from part b, estimate the stopping distance for a car travelling at these speeds. i 50 km/h ii 120 km/h
Exercise 1.5G Challenge
Answers: pXXX
20 The formula for the total stopping distance is given by Stopping distance = reaction-time distance + breaking distance.
The average reaction time for a driver is 2 seconds. The breaking distance formula is given by d = 0.01v 2 New South Wales Road and Maritimes Services claim that “If you are driving at 50 km/h, it wil l take you about 37 metres to stop.” Compare this claim with the result of the stopping distance formula above.
Related resources
1.5 Quick quiz
Investigation: Travelling at constant speed
Lesson 1.6 Blood alcohol content calculations
Learning intentions
By the end of this lesson, you will be able to ...
→ solve problems related to blood alcohol content.
Blood alcohol content
• Blood alcohol content (BAC) is a measure of the concentration of alcohol in a person’s blood. It is expressed as a percentage mass per unit of volume.
→ A person with a BAC of 0.02 (%) has 0.02 100 g of alcohol in every millilitre of their blood. This is equivalent to 0.02 g/100 mL or 20 mg/100 mL
• A BAC measure can be estimated using tables, formulas and on-line calculators; but it is very important to remember that these are only approximations because they are based on average values and do not apply equally to everyone.
• A person’s BAC reading is affected by factors such as whether the person is male or female, how much the person drank, the length of time that the person spent drinking and the length of time since his or her last drink, the person’s weight, whether the person is physically fit, the state of the person’s liver, whether the person is a regular drinker, or the person’s mood at the time.
• The only way to measure a person’s BAC accurately is with an approved breath analysing unit known as a ‘breathalyser’.
Standard drinks
• An estimate of a person’s BAC can be determined by counting the number of standard drinks consumed. A standard drink is the quantity of a beverage that contains 10 g of alcohol. A standard drink always contains the same amount of alcohol irrespective of the container size or the type of drink (beer, wine or spirits).
• The number of standard drinks in a container can be calculated using the formula:
N = 0.789 × V × A where:
N = number of standard drinks
V = the capacity of the container in litres
A = percentage of alcohol (% alc/vol) in the drink.
→ The percentage of alcohol in the drink is stated on the container.
→ 0.789 is the specific gravity of ethyl alcohol, the type of alcohol used in beverages.
Estimating blood alcohol content
• An estimate of a person’s BAC can be found using these formulas:
where:
N = number of standard drinks consumed
H = number of hours spent drinking
M = person’s mass in kg.
→ One standard drink per hour will raise BAC by between 0.01% and 0.03%
• A person’s BAC will increase at a greater rate if the person:
→ is female
→ is drinking highly carbonated drinks
→ has a low body weight
→ is unfit
→ has not eaten recently
→ has an unhealthy liver.
Reducing blood alcohol content
• After drinking ceases, the only thing that will reduce BAC is the passing of time.
→ Drinking coffee, exercising, taking a cold shower or inducing vomiting will not reduce BAC.
→ Alcohol is eliminated from the body by the liver at a rate between 4 g/h and 12 g/h, or at an average of 7.5 g/h or 0.75 standard drinks per hour (since a standard drink contains 10 g of alcohol), and can vary considerably depending on the person’s health. This means that it can take the liver more than an hour to eliminate one standard drink.
• The liver breaks down alcohol at an average rate of 0.75 standard drinks per hour.
• A formula that can be used to calculate the time it takes for BAC to fall to zero is:
T = BAC 0.015 , where T is the number of hours the drinker must wait.
Worked example 1.6A Calculating the number of standard drinks
Calculate the number of standard drinks in a 150 mL glass of red wine, correct to one decimal place, given that the alcohol content of the wine is 14.5% alc/vol.
Think Write
Convert the capacity of the glass to litres using:
1 L = 1000 mL
The alcohol content is 14.5% alc/vol, so A = 14.5.
Substitute these values into the formula V = 0.789 × V × A
Simplify and write the result correct to one decimal place.
Worked example video: Calculating the number of standard drinks
V = 150 mL = 150 × 1 1000 L = 0.15 L
N = 0.789 × V × A = 0.789 × 0.15 × 14.5
≈ 1.7
Exercise 1.6A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Remember to convert the capacity of the container to litres before substituting it into the formula.
1 Complete the following to calculate the number of standard drinks in these beers correct to one decimal place.
a A 375 mL stubby of full-strength beer with an alcohol content of 4.8% alc/vol
N = 0.789 × V × A = 0.789 × __ × 4.8 = __
b A 375 mL stubby of light beer with an alcohol content of 2.7% alc/vol
N = 0.789 × V × A = 0.789 × 0.375 × __ = __
2 Calculate, correct to one decimal place, the number of standard drinks in:
a a 120 mL glass of wine with an alcohol content of 12% alc/vol
b a 90 mL glass of fortified wine with an alcohol content of 16.5% alc/vol
c a 375 mL can of bourbon and coke with an alcohol content of 6% alc/vol
d a 750 mL bottle of white wine with an alcohol content of 11.5% alc/vol
e a 275 mL bottle of vodka and orange with an alcohol content of 5% alc/vol
f a six-pack (6 × 330 mL) of full strength beer with an alcohol content of 4.9% alc/vol.
Worked example 1.6B Calculating blood alcohol content (BAC)
Calculate, correct to two decimal places, the BAC of:
a a 78 kg male who has consumed five standard dr inks in 3 hours
b a 46 kg female who has consumed four standard drinks in 4 hou rs.
Think Write
a Substitute N = 5, H = 3 and M = 78 into the formula
BAC male = 10N − 7.5H 6.8M .
Simplify and write the result correct to two decimal places.
b Substitute N = 4, H = 4 and M = 46 into the formula
BAC female = 10N − 7.5H 5.5M
Simplify and write the result correct to two decimal places.
Worked example video: Calculating blood alcohol content (BAC)
If he has standard drinks, he will have BAC = 0.05.
To have BAC < 0.05, he ca n have up to standard drinks.
6 A 56 kg wom an wants a BAC that is less than 0.05. How m any drinks can she consume in 3 hou rs?
7 A rule of thumb can be used by the holder of a full drivers’ licence to stay under the 0.05 lega l limit in NSW:
For males: no more than two standard drinks in the first hour and one standard drink per hour after that.
For females: no more than one standard drink per hour.
Use this rule of thumb to calculate the maximum number of standard drinks that can be consumed by a person who wants to stay under 0.05 BAC i f the person is:
a a male and drinking for 4 hours
b a male and drinking for 6 hours
c a female and drinking for 4 hours
d a female and drinking for 6 hours.
8 Use the rule of thumb from question 7 to calculate the answers to questions 5 and 6 .
Worked example 1.6C Calculating
time needed to eliminate one standard drink
If a person’s liver can break down alcohol at the rate of 5 g/h, how long will it take for this person’s body to eliminate one standard drink? Write your answer correct to the nearest minute.
Think Write
Let T be the amount of time required. One standard drink contains 10 g of alcohol. Write the given rate as a fraction. Find an equivalent rate with 10 g as the numerator. Divide 10 by 5 to identify the value required to find the equivalent rate. Multiply the numerator and denominator by this value to find the equivalent rate. The time taken is the denominator of the equivalent rate.
Worked example video: Calculating time needed to eliminate one standard drink
Exercise 1.6C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
DRAFT
• If you are unsure where to start with a rate question, write the rate as a fraction and include the units in the numerator and denominator. You can then work with the rate by finding equivalent fractions with the same units.
→ If the rate at which the liver eliminates alcohol is given in g/h, then the time in hours needed to eliminate one standard drink is 10 rate of elimination hours.
→ Check you are using the correct formula to answer the question. If asked to find the amount of time taken for a given BAC to reduce to zero, use T = BAC 0.015
5 Calculate the time it takes different people’s bodies to eliminate one standard drink if their livers break down alcohol at these rates.
a 5 g/h b 10 g/h c 4 g/h d 12 g/h e 7.5 g/h
6 Complete the following to calculate how long a person must wait for BAC to drop to zero from 0.05% T = BAC 0.015 = □ 0.015 = __ h = __ h __ min
7 Calculate how long a person must wait for BAC to drop to zero from these levels.
a 0.04% b 0.02% c 0.06% d 0.035%
Exercise 1.6D Problem solving and reasoning
Answers: pXXX
8 A zero BAC is a requirement of NSW law for all learner and provisional drivers.
a Calculate the BAC for an 80 kg male and a 52 kg female, both with provisional licences, who consume four standard drinks in 3 hours.
b How long would the two people described in part a need to wait before they could legally drive a motor vehicle?
9 Damien and Nicole go to a party and start drinking at 8 : 00 pm. Dam ien drinks eight schooners of fullstrength beer (12 sta ndard drinks) over the next 5 hou rs. Nicole has six mixed drinks (9 sta ndard drinks) in the same time. Damien has a mass of 86.6 kg and Nicole’s mass is 56.1 kg
a Calculate the BAC of both Damien and Nicole at 1 : 00 am.
b At what time will they be able to legally drive if they both have provisional licences?
10 Ben goes to a party and consumes two stubbies (375 mL) of fu ll-strength beer (4.9% alc/vol) in the first hour and one stubby per hour for the next 3 hou rs.
a Calculate the number of standard drinks he has consumed.
b Use the formula to calculate his BAC if his mass is 72 kg
c How long will it be before Ben’s BAC drops to zero?
Exercise 1.6E Challenge
Answers: pXXX
11 Rebecca goes out with friends for drinks and starts drinking at 8 : 30 pm. Over the next 3 hou rs, she consumes 4.8 sta ndard drinks.
a She stops drinking at 11 : 30 pm. Calculate her BAC at this time.
b She needs to wait for her BAC to get below 0.05 before she drives home. At what time should her BAC be below 0.05?
Lesson 1.7
Medication calculations
Learning intentions
By the end of this lesson, you will be able to ...
→ calculate medical dosages using Fried’s, Young’s and Clark’s formulas.
Dosage calculations
• This section examines dosages of various medications. Some terms are defined here.
→ The dose is the amount of drug taken at any one time.
→ The dosage regimen is the frequency at which the drug doses are given.
→ The total daily dose is calculated from the dose and the number of times the dose is taken.
→ The dosage form is the physical form of a dose of the drug. Common dosage forms include tablets, capsules, creams, ointments, aerosols and patches.
→ The optimal dosage is the dosage that gives the desired effect with minimal side effects.
• Failing to adhere to the instructions for self-administered medications can have potentially serious consequences. It is important to understand both the maximum frequency and total daily dosage restrictions.
• When administering medication to children, doses must be adjusted appropriately. There are three different formulas that can be used to calculate dosage for children, each depending on the age of the child.
→ Fried’s formula for children 1 to 2 year s old is:
Dosage = age (in months) × adult dosage 150 = mA 150
where:
D = infant dosage
m = age of infant in months
A = adult dosage.
→ Young’s formula for children 1 to 12 year s old is:
Dosage = age of child (in years) × adult dosage age of child (in years) + 12 = yA y + 12
where:
y = age of child in years
A = adult dosage.
→ Clark’s formula for children of any age is:
Dosage = child"s weight (in kilograms) × adult dosage 70 = kA 70
where:
k = mass of child in kilograms
A = adult dosage.
• Some medications are administered over a period of time, for example, via intravenous fluid. In these contexts, determining an appropriate dosage regimen requires an understanding of rates and accurate conversion of units.
Worked example 1.7A Understanding maximum dosages
The dosage instructions on a packet of painkillers say that a person 12 yea rs or older can take 2 tablets every 4 to 6 hou rs, but only up to a maximum of 8 tablets in 24 hou rs.
An adult plans to take two tablets every 4 hou rs for 24 hou rs.
a How many tablets would that person take over 24 hours? Why shouldn’t they do this?
b How many doses of two tablets can be taken over 24 hou rs?
Think Write
a Let N be the number of tablets. Write information provided as a rate in fraction form. Find an equivalent rate with 24 hou rs as the denominator. Divide 24 by 4 to determine there are six 4-hour intervals in a 24-hour period. Multiply the numerator and denominator by 6 to find the equivalent rate.
The number of tablets taken over 24 hou rs is the numerator.
Compare the value of N with the maximum daily dose.
Since the maximum daily dose is 8 tablets, this dosage regimen exceeds the maximum.
b The maximum daily dose is 8, which can be divided into 4 doses of 2 tablets. Doses = 8 2 = 4
Worked example video: Understanding maximum dosages
Exercise 1.7A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Watch out for maximum dosage frequencies that would exceed the maximum daily dosage if followed throughout the day.
1 The dosage for a particular painkiller is as follows.
Age: 7 to 12 years: 1 2 to 1 tablet every 4 to 6 hour s (maximum 4 tablets in 24 hour s)
Age: 12 years to adult: 1 to 2 tablets every 4 to 6 hour s (maximum 8 tablets in 24 hour s)
DRAFT
a How many tablets can an adult take in one dose?
b An adult plans to take two tablets every 6 hours for 24 hours.
i How many tablets would that person take over the 24 hours?
ii How many doses of one tablet could they take over 24 hours?
c A child takes 1 2 a tablet every 4 hours for 24 hours. Has the maximum dosage been exceeded? Explain.
2 For a very strong painkiller, dosage for adults and children from 12 yea rs is two caplets, then 1 to 2 caplets every 4 to 6 hou rs as necessary. (Maximum 6 caplets in 24 hou rs.)
a An adult takes two caplets now and then two more after 4 hours. How many more caplets can that person take in that 24-hour period?
b Is two caplets initially, two more after 4 hours and two more after 6 hours, then no more, an acceptable dosage? Explain your answer.
Worked example 1.7B Using Fried’s formula to calculate medication dosage
The adult dose of a medication is 40 mL. Use Fried’s formula for children 1 to 2 yea rs old to calculate the dosage for a 20-month-old child.
Think Write
Use the formula D = mA 150 and substitute m = 20 months and A = 40 mL.
Dose = 20 × 40 150 = 5.3 mL
Worked example video: Using Fried’s formula to calculate medication dosage
Exercise 1.7B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Recall that Fried’s formula for children 1 to 2 year s old is:
Dosage = age (in months) × adult dosage 150 or D = mA 150 where D = infant dosage, m = age of infant in months and A = adult dosage.
• Ensure that the units are correct for the formula; that is, age is in months.
3 Use Fried’s formula to calculate each child’s dosage below.
a Adult dose of 50 mL, child’s age 15 months
b Adult dose of 40 mL, child’s age 21 months
c Adult dose of 30 mL, child’s age 18 months
d Adult dose of 50 mL, child’s age 13 months
e Adult dose of 100 mL, child’s age 17 months
f Adult dose of 80 mL, child’s age 23 months
4 Use Fried’s formula to calculate the adult dosage in these situations.
a A child aged 17 months is given a dosage of 6 mL
b A child aged 11 months is given a dosage of 7 mL
Worked example 1.7C Using Young’s formula to calculate medication dosage
Use Young’s formula to calculate the dosage for a 5 1 2 -year-old child if the adult dose is 60 mL
Think Write
Use the formula D = yA y + 12 where y = 5.5 and A = 60 mL
Worked example video: Using Young’s formula to calculate medication dosage
Exercise 1.7C Understanding, fluency and communicating
Helpful hints
• Recall that Young’s formula for children 1 to 12 year s old is Dosage = age of child (in years) × adult dosage age of child (in years) + 12 or D = yA y + 12 where y = age of child in years and A = adult dosage.
• Check the units before substituting. Some formulas use age in years, others use age in months.
5 Use Young’s formula to calculate each child’s dosage, to the nearest millilitre.
a Adult dose of 50 mL, child’s age 6 year s
b Adult dose of 40 mL, child’s age 8 year s
c Adult dose of 80 mL, child’s age 4.5 year s
d Adult dose of 20 mL, child’s age 7.5 year s
e Adult dose of 100 mL, child’s age 6.2 year s
f Adult dose of 75 mL, child’s age 8.4 year s
6 Use Young’s formula to calculate the adult dose in these situations.
a The dosage for an 8-year-old child is 10 mL.
b The dosage for a 6-year-old child is 5 mL
Worked example 1.7D Using Clark’s formula to calculate medication dosage
Use Clark’s formula to calculate the dosage for a child weighing 19 kg. The adult dose is 50 mL
Think Write
Use the formula D = kA 70 where k = 19 and A = 50 Dose = 19 × 50 70 ≈ 14 mL
Worked example video: Using Clark’s formula to calculate medication dosage
Exercise 1.7D Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Recall that Clark’s formula for children of any age is:
Dosage = child’s weight (in kilograms) × adult dosage
70 or D = kA
70 where k = mass of child in kilograms and A = adult dosage.
• Check units before substituting. The formula uses 70 kg as the average adult weight.
7 Use Clark’s formula to calculate each child’s dosage, to the nearest millilitre.
a Adult dosage of 18 mL, child’s weight 40 kg
c Adult dosage of 51 mL, child’s weight 80 kg
e Adult dosage of 32 mL, child’s weight 100 kg
b Adult dosage of 27 mL, child’s weight 60 kg
d Adult dosage of 39 mL, child’s weight 75 kg
f Adult dosage of 40 mL, child’s weight 35 kg
8 Use Clark’s formula to calculate the adult dose in these situations.
a A 35 kg child has a dose of 18 mL
b A 25 kg child has a dose of 17 mL
9 Use Clark’s formula to calculate the weight of a child receiving a dose of 40 mL, given that the adult dose is 140 mL.
Exercise 1.7E Problem solving and reasoning
Answers: pXXX
10 A sick child requires medication. The adult dose of the medication is 20 mL. The child is 16 months old and has a mass of 7.9 kg.
a Find the dosage using Fried’s rule.
c Find the dosage using Clark’s rule.
b Find the dosage using Young’s rule.
d How much would you give the child? Explain.
11 Repeat question 15 for a 20-month-old child weighing 12 kg, given that the adult dose of the medication is 30 mL.
12 A child is 10 yea rs of age and requires a dose of a medicine for which the adult dosage would be 50 mL. The dosage is calculated using Fried’s rule, without realising that rule is used for children under 2 yea rs only. A dose could be an overdose if it is double the correct dose.
a Calculate the dosage according to Fried’s rule. b Calculate the correct dosage using Young’s rule.
c If the entire amount calculated using Fried’s rule is taken at the one time, has the child received an overdose? Explain.
d The dosage is given through a drip at 20 drops per minute with 15 drops per millilitre. At what time should the drip be stopped to result in the correct dosage being given?
Exercise 1.7F Challenge
Answers: pXXX
13 For a certain medication, the adult dosage is 13 mL. Usi ng Young’s formula, calculate the age of a child who would receive a dosage of 5 mL
Related resources
1.7 Quick quiz
Lesson 1.8
Solving simple quadratic equations
Learning intentions
By the end of this lesson, you will be able to ...
→ solve simple quadratic equations and assess whether the solution is reasonable.
Simple quadratic equations
• A simple quadratic equation is an equation of the form y = ax 2 + c
• When y = 0, a simple quadratic equation can be solved for x using inverse operations.
→ The inverse operation to squaring is taking the square root.
• A quadratic equation can have zero, one or two solutions.
→ The solutions to 0 = ax 2 + c are x = √ c a and x = −√ c a .
→ If c a > 0, then there are two solutions. These are often written as x = ± √ c a
→ If c a = 0, then √ c a = −√ c a , so the only solution is x = 0.
→ If c a < 0, then there are no real solutions.
• Square roots of fractions can be simplified in two ways.
→ The rule √ a b = √a √b , where both a and b are positive values, can be used to simplify the square root of a fraction.
→ Square roots of fractions can also be simplified by first writing the fraction in its simplest form before taking the square root.
• In some contexts, a negative solution to an equation may not make sense. For example, measurements like distances, lengths and durations of time are usually positive values.
Worked example 1.8A Solving
simple quadratic equations (exact solutions)
Solve the following simple quadratic equations.
a 3x 2 = 27 b −4x 2 = −25 c
Part Think Write
a Divide both sides of the equation by 3 and simplify the result. Take the square root of both sides of the equation to solve for x As 9 > 0, there are two solutions to the equation. 3x 2 = 27 x 2 = 27 3 x 2 = 9 x = ± √9 x = ± 3
Part Think Write
b Divide both sides of the equation by −4 and simplify the result. Take the square root of both sides of the equation to solve for x As 25 4 > 0, there are two solutions to the equation.
Use the rule √ a b = √a √b to find the square root of the fraction and simplify the result.
c Add 0.08 to both sides of the equation.
Multiply both sides of the equation by 2 and simplify the result. Take the square root of both sides of the equation to solve for x As 0.16 > 0, there are two solutions to the equation.
Worked example video: Solving simple quadratic equations (exact solutions)
Exercise 1.8A Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Make sure to only take the square root of a positive number (or zero) when solving simple quadratic equations.
1 Solve the following simple quadratic equations.
2 Solve the following simple quadratic equations.
4x 2 = 81
−25x 2 = −121
3 Solve the following simple quadratic equations.
x 2 = 0.01
4 Solve the following simple quadratic equations.
2x 2 − 1 = 31
5x 2 − 2000 = 0
x 2 2 − 3.125 = 0
Worked example 1.8B Solving simple quadratic equations (approximations)
Solve the following simple quadratic equations, correct to two decimal places.
a 2x 2 = 3
b 0.2x 2 − 1.7 = 0
Part Think Write
a Divide both sides of the equation by 2
Take the square root of both sides of the equation. As 3 2 > 0, there are two solutions to the equation.
Use a calculator to write the solutions correct to two decimal places.
b Add 1.7 to both sides of the equation.
Divide both sides of the equation by 0.2 and simplify the result.
Take the square root of both sides of the equation. As 8.5 > 0, there are two solutions to the equation.
Use a calculator to write the solutions correct to two decimal places.
Worked example video: Solving simple quadratic equations (approximations)
Exercise 1.8B Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Pay attention to the order of operations when typing more complicated expressions into a calculator. Check that the answer you get makes sense.
5 Solve the following simple quadratic equations, correct to two decimal places.
a 3x 2 = 2
b 5x 2 = 7
c 4x 2 − 11 = 0 d 5x 2 − 17 = 0
6 Solve the following simple quadratic equations.
a x 2 + 1.5 = 2.9
b 1.7x 2 − 3.2 = 0
7 Solve the following equations.
a a 2 2.7 − 7.2 = −4.5
b 1 3 (p 2 − 12) = −4
DRAFT
c 0.3x 2 + 5.1 = 9.2
d 2.9x 2 − 10 = 0
c 2M 2 + 0.0008 = 0.008
d 3k 2 − 9 11 = 0
Worked example 1.8C Applications of simple quadratic equations
The formula for the surface area of a cube is SA = 6s 2, where s is the side length of the cube. Determine the side length of a cube with a surface area of 150 cm 2 . s
Think Write
Substitute SA = 150 cm 2 into the formula for the surface area of a cube.
Divide both sides of the equation by 6 and simplify the result.
Take the square root of both sides of the equation. As 25 > 0, there are two solutions.
The side length of a cube must be a positive value. Take only the positive solution to the equation.
Worked example video: Applications of simple quadratic equations
Exercise 1.8C Understanding, fluency and communicating
Answers: pXXX
Helpful hints
• Make sure the units of measurement in your solutions match what is required in the question.
• Check that your final answer makes sense in the context of the question. For example, a cube cannot have a side length of −5 cm!
8 The formula for the surface area of a cube is SA = 6s 2, where s is the side length of the cube. Determine the side length of a cube with a surface area of 96 cm 2
DRAFT
9 The braking distance in metres of a car travelling at v km/h in good conditions is d = 0.01v 2. What is the speed of a car which has a braking distance of 42.25 m?
10 The braking distance in metres of a car travelling at v km/h in poor conditions is d = 0.014v 2. What is the speed of a car which has a braking distance of 47.096 m?
11 Two cars are travelling in good conditions at different speeds. One car has a braking distance of 49 m and t he other has a braking distance of 56.25 m. What is the difference between the speeds of the two cars in km/h?
Exercise 1.8D Problem solving and reasoning
Answers: pXXX
12 Consider the equation y = 2x 2 + 3
a Substitute y = 2 into the equation and solve for x
b Substitute x = 2 into the equation and solve for y
c How many solutions are there to parts a and b? Explain why this is the case.
13 Consider the equation y = −3x 2 − 4
a For which value of y is there only one solution to the equation?
b Find a value of y for which there is no solution to the equation.
14 Which of the following equations have no solutions? a x 2 + 1 = 0
2x 2 = 9
15 Write a simple quadratic equation that has no solutions and then show that it has no solutions.
Exercise 1.8E Challenge
Answers: pXXX
16 Solve the following equation after substituting p = 16, n = 5 and k = 7. Write your answer(s) correct to two decimal places. p = n + 12kt 2
Related resources
1.8 Quick quiz
DRAFT
Worksheet: Solving simple quadratic equations
Using spreadsheets
Syllabus link
Use a spreadsheet to perform calculations involving formulas (MST-11-01)
Formulas and equations
See Oxford Digital for sample spreadsheets for questions in this lesson.
1 Use a spreadsheet program to solve the equation 3x − 7 = 9 − x, given that the solution is an integer. The spreadsheet needs to have three columns labelled x, 3x − 7 and 9 − x
In cell A1 enter x, in cell B1 enter 3x − 7, in cell C1 enter 9 − x, in cell A2 enter 0, in cell B2 enter = A2*3-7 and in cell C2 enter = 9-A2.
Use the fill down command to find the values for each side of the equation. The answer is the value of x that gives the same value in columns B and C
2 Change the spreadsheet in question 1 to solve each of these equations with positive integer solutions. a 5x − 3 = 53 − 2
3 Use a spreadsheet to solve these equations with negative integer solutions. a 4x − 3 = 13 + 6x b 7x − 3 = −25 + 5x
4 Explain how to modify the spreadsheet used in questions 2 and 3 to solve equations that do not have integer solutions.
5 Solve these equations using a spreadsheet.
a 3x − 7 = 6 − 2x b 5x + 23 = 2x − 8 c 8 − 7x = 8x + 59
6 Create a spreadsheet to show the blood alcohol content for a 75 kg male a nd a 55 kg fem ale, where N is the number of standard drinks consumed compared with the number of hours spent drinking. Investigate what the BAC would be for a given number of standard drinks over different periods of time. Remember it is only an approximation!
7 Create a spreadsheet to calculate the dosage of a particular medicine for children under 12 yea rs of age. Fried’s rule is applicable for children under 2 yea rs, Young’s formula is for children 1 to 12 yea rs and Clark’s formula is for children of any age. Set your spreadsheet up so that when you enter the three variables for a child the spreadsheet automatically calculates their dosage.
1 Fill in the blanks using words from the list above.
• solve
• standard drink
• subject of the formula
• term (algebra)
• total stopping distance
• variable (algebra)
a Any letter that appears in a mathematical expression is called a .
b To an equation means to find a value for a that makes the equation true. The value found is a to the equation.
2 Write the meaning of the word formula in a sentence including the words equation and variable.
3 Give three examples of a linear equation and three examples of a quadratic equation
4 Label one part of the following equation that is an example of an expression and label three different parts that are examples of a term
5x − 2y + 1 = 7y 2 + x
5 Fill in the blanks using words from the list above.
a Australian adults who consume alcohol responsibly will often take note of how many they consume. This way, they can make a reasonable estimate of their before deciding to engage in risky behaviours, such as driving a car.
b To evaluate a driver’s ability to avoid an accident by coming to a complete stop, we are interested in their , which is directly influenced by their . The total distance is found by adding together the and .
Review questions 1.9B Multiple-choice questions
Answers: pXXX
1 Let a = 7 and b = −3. What is the value of a 2 − 5b 8 ?
2 What is the value of s = ut + 1 2 at 2 if u = 0, t = 5 and a = 10?
3 The volume of a cone with h = 10 and r = 3 is calculated using V = 1 3 πr 2h. Which value is closest to the volume?
A 31
B 94
C 314
D 30
4 What is the solution to p − 16 = 42?
A p = 26
B p = 36
C p = 48
D p = 58
5 What is the solution to 11x − 10 = 25?
A x = 25
B x = 3 2 11
C x = 11 35
D x = 15 11
6 Which equation has solution x = −3?
A 3x − 7 = 2
B 4x − 3 = 15
C x + 5 = 8
D 5x + 2 = −13
7 Which equation does not have x = 4 as a solution?
A x − 4 = 0
B x 4 = 16
C 4x = 16
D x + 5 = 9
8 What is the solution to 3a + 4 5 = 2 ?
A a = 2
B a = 3
C a = 4
D a = 5
9 What is the solution to 7x − 3 = 4x + 9?
A x = 2
B x = 3
C x = 4
D x = 5
11 The formula V = IR − E is rearranged to make R the subject. What is the new formula?
A R = V + E I
B R = V − I E
C R = E VI
D R = E − V I
12 A car travels 280 kilometres in 3 hours and 25 minutes. What is its average speed?
A 86 km/h
B 82 km/h
C 93 km/h
D 42 km/h
13 What is 70 km/h expressed in m/s?
A 1.2 m/s
B 0.02 m/s
C 1167 m/s
D 19.4 m/s
14 What is the distance a car travels in 2.8 seconds at 80 km/h?
DRAFT
10 Using the formula V = IR − E, what is the value of I when V = 13, R = 4 and E = 7?
A 1 1 2
B 1 4
C 5
D 2 3
A 6.2 m
B 7.9 m
C 62.2 m
D 373.3 m
15 Paul goes to a party and consumes three stubbies (375 mL) of full-strength beer (4.9% alc/vol) in the first hour and one stubby per hour for the next 4 hours. Approximately, how many standard drinks has he consumed?
A 1.4
B 4.3
C 5.8
D 10.1
16 A 75 kg woman consumes seven standard drinks in 4 hours. What is her BAC?
A 0.078
C 0.010
B 0.097
D 0.012
17 The adult dose of a medication is 50 mL.
Using Fried’s formula below, what is the dosage for a 1-year-old child?
Child dose = age (in months) × adult dose
150
A 4 mL
B 0.3 mL
C 3 mL
D 0.4 mL
18 What are the solutions to the quadratic equation
4x 2 − 9 = 7?
A x = 2
B x = 3 2
C x = ± 2
D x = ± √ 1 2
Review questions 1.9C Set 1
Answers: pXXX
1 If x = 4 and y = 6, find the value of 3x − 5y
19 How many solutions does the quadratic equation
2x 2 − 3 = −4 have?
A One solution, since x = √ −1 2
B No solutions, since 1 2 < 0
C One solution, since x = −√ 1 2
D No solutions, since 7 2 < 0
2 The volume, V, of a sphere is found using the formula V = 4 3 πr 3, where r is the radius. Find the volume of a sphere with radius:
a 7 cm
b 3.5 cm
c 0.6 m
3 The formula for converting temperature measurements from degrees Celsius, C, to deg rees Fahrenheit, F, is F = 9 5 C + 32. Find F when:
a C = 200 °
4 Solve the following equations.
a d − 8 = 36
d 11x = 55
b C = 20 °
b x + 9 = 20
e x 2 − 3 = 46
c C = 55 °
c x 8 = 72
f 3x − 12 = 60 g 10 + 4n 2 = 74
12 − 6c = 85
5 Use the formula E = 1 2 mv 2 to find the positive value of v when m = 50 and E = 10 000
6 Make x the subject of each formula below.
a 3x = t
b y = mx − 7
8 A train averages 84 km/h for 2 hours and 36 minutes. How far does it travel?
3(2x + 1) = 4
c a = x + d c
7 For a car travelling on a slippery road, the formula for braking distance becomes d = 0.015v 2. What is the braking distance of a car travelling at 100 km/h on this road?
9 Calculate how long an average person must wait for BAC to drop to zero from 0.07%
10 Use Fried’s formula to calculate the dosage of medicine for a 9-month-old child if the adult dose is 60 mL
11 Use Young’s formula to calculate the dosage of medicine for a 6 1 2 -year-old child if the adult dose is 45 mL.
Review questions 1.9D Set 2
Answers: pXXX
1 If x = −2 and y = 3, find the value of 5x − 7y.
2 The surface area, A, of a sphere is found using the formula A = 4πr 2, where r is the radius. Find the surface area of a sphere with radius:
a 4 cm
DRAFT
b 7.5 cm
c 0.8 m
3 The formula for converting temperature measurements from degrees Fahrenheit, F, to deg rees Celsius, C, is C = 5 9 (F − 32). Find C when:
a F = 248 °
4 Solve the following equations.
a d − 9 = 23
d x 2 = 121
g 12 − 9c = 43
b F = 50 °
b x + 4 = 65
e 3x − 7 = 30
h 7(2x + 5) = 4
c F = 32 °
c x 3 = 5
f 3x 2 − 12 = 15
i 4x − 2 = x + 9
5 Use the area formula A = 1 2 bh to find the length b, given that A = 72 and h = b
6 Make x the subject of each formula below.
a x + y = m
b n = k − 3x
7 A car travels 252 km in 4 hours and 19 minutes. Calculate its average speed.
c m = x k − 4
8 Calculate the BAC of a 98 kg male who has consumed six standard drinks in 4 hours.
9 Use Young’s formula to calculate the dosage of medicine for a 7 1 2 -year-old child if the adult dose is 25 mL
10 Use Clark’s formula to calculate the dosage of medicine for a child weighing 15 kg. The adult dose is 30 mL
Review questions 1.9E Set 3
Answers: pXXX
1 Given x = −5 and y = 8, find the value of 4x 2 − 6y
2 The surface area, A, of a cylinder is found using the formula A = 2πr(r + h), where r is the radius and h is the height. Find the surface area of a cylinder with radius 15 cm and height 10 cm
3 The time T seconds for a pendulum of length L m to swing back and forth once is give by the formula T = 2π
√ L g where g ≈ 10 m/s 2. Find how long it takes for a pendulum of length 2.5 m to swing back and for th once.
4 Solve the following equations.
a d − 9 = 11
d 11x = 77
g 14 + 3n = 23
b x + 4 = 14
e 3x − 8 = 12
h 23 − 3c 2 = 11
c x 3 = 2
f 7x − 12 = 19
i 2(4x + 5) = 12
5 Use the formula T = a + (n − 1)d to find n, given that T = 41, a = 3 and d = 2
6 Make x the subject of each formula below.
a x y = t
b y = k − mx 2
c x − t w = y
7 Find the total stopping distance for a car travelling at 90 km/h in good conditions. Assume a reaction time of 2.5 seconds and a braking distance of d = 0.01v 2 .
8 Calculate the number of standard drinks in a 120 mL glass of red wine, g iven that the alcohol content of the wine is 14.6% alc/vol.
9 Calculate the BAC for:
a a 76 kg male who has consumed five standard dr inks in 3 hours
b a 52 kg female who has consumed four standard drinks in 4 hours.
10 Calculate how long an average person must wait for BAC to drop to 0 from 0.06%
11 A zero BAC is a requirement of NSW law for all learner and provisional drivers.
a Calculate the BAC for a 75 kg male and a 53 kg female, both with provisional licences, who consume four standard drinks in 3 hours.
b How long would the two people have to wait before they could legally drive a motor vehicle?
DRAFT
12 Use Clark’s formula to calculate the dosage of medicine for a child weighing 38 kg , given the adult dose is 35 mL
Review questions 1.9F Set 4
Answers: pXXX
1 Given x = 12 and y = −4, find the value of 6x 2 − 11y
2 The volume, V, of a cylinder is found using the formula V = πr 2h, where r is the radius and h is the height. Find the volume of a cylinder with radius 12 cm and height 25 cm
3 A formula for calculating the bend allowance, in millimetres, of sheet metal is B = 2π(R + T 2 ) × A 360 where B is the bend allowance, T is the thickness in millimetres, A is the number of degrees in the angle of bend, and R is the radius of curvature in millimetres. Find B when T = 1.5, R = 4 and A = 113
4 Solve the following equations.
a d − 5 = 11
d 7x = 77
b x + 7 = 14
e 3x − 4 = 92
g 11 + 7n = 53 h 27 − 9c = 31
5 Use the formula b = m h 2 to find positive h if b = 31.25 and m = 80.
6 Use the formula a = 2Rn n + 1 to find the value of R when a = 95 and n = 19.
7 If a cyclist can average 15 km/h, how long will it take her to travel 28 km?
8 Convert the following.
a 55 km/h to m/s
c x 6 = 2
f 8x 2 − 13 = 11
i 4(3x + 5) = 13
b 21.4 m/s to km/h
9 Calculate the reaction-time distance for a car travelling at 60 km/h. Assume a reaction time of 2.5 seconds.
10 Use the formula d = 0.01v 2 to calculate the braking distance for a car travelling at 80 km/h in good conditions.
11 Find the total stopping distance for a car travelling at 80 km/h in good conditions. Use the formula d = 0.7v + 0.01v 2, where d is the stopping distance in metres and v is the speed in km/h
12 For a driver under the influence of alcohol and driving in poor road conditions, the formula for the stopping distance of a car becomes d = 1.1v + 0.018v 2
a Calculate the stopping distance to the nearest metre to complete the table for this driver.
Stopping distance (m)
b Use the table from part a to draw a graph of the relationship between speed and stopping distance.
c From the graph drawn for part b estimate the stopping distance if the car is travelling: i 50 km/h ii 70 km/h iii 110 km/h
13 Use Young’s formula to calculate the dosage of medicine for a 10-year-old child. The adult dose is 20 mL
DRAFT
Review questions 1.9G Practice examination question
Answers: pXXX
1 Answer the following questions.
a Solve x 3 = −3 (1 mark)
b Find the value of √10a + 2b when a = 5 and b = 7. (2 marks)
c The body mass index, b, is used to determine whether a person’s mass is within the acceptable range of 21 < b < 25. To calculate her body mass index, Stella used the formula b = m h 2 , where m = mass in kilograms and h = height in metres.
Here are her calculations.
b = m h 2
When m = 55 kg and h = 168 cm:
b = 55 168 2 = 0.001 95
i Explain the error in her calculations. (1 mark)
ii Calculate the correct value of b. (1 mark)
iii Is Stella in the healthy mass range? Explain. (1 mark)
d George’s solution to the equation 6 − 5x = 8 is shown below.
6 − 5x = 8
6 − 5x + 6 = 8 + 6 −5x = 14
x = 8
He has made some errors. Identify the errors and find the correct solution. (2 marks)
e The total stopping distance, d metres, of a car travelling at v km/h under good conditions is given by the formula d = 0.7v + 0.01v 2
i Calculate the stopping distance of a car travelling at 60 km/h (1 mark)
ii If the speed of the car is increased by 10 km/h, what is the increase in the stopping distance? (2 marks)
f Harry goes to a party and drinks seven standard drinks over 4 hours.
i Use the formula BAC male = 10N − 7.5H 6.8M to calculate Harry’s blood alcohol content after 4 hours if his mass is 78 kg. (2 marks)
ii Use the formula T = BAC 0.015 to find how long it will take for his BAC to drop to zero. (1 mark)
g Use Young’s formula below to calculate the dosage of medicine for a 4 1 2 -year-old child, given the adult dose is 100 mL.
Dosage for children 1 to 12 years = age of child (in years) × adult dosage age of child (in years) + 12 (1 mark)
TOTAL: 15 marks
Checklist
Now that you have completed this module, reflect on your ability to do the following.
Checklist: Formulas and equations
I can do this
⃞ substitute numbers into expressions, equations and formulas
⃞ evaluate the value of the subject of a formula given the value of other pronumerals in the formula
⃞ solve linear equations
⃞ solve equations by substitution into formulas
⃞ change the subject of a formula that is a linear equation
⃞ solve problems related to speed, distance and time
⃞ change units when solving problems related to speed, distance and time
⃞ calculate the distance a motor vehicle travels in the time it takes to bring it to a stop
⃞ calculate the number of standard drinks in a serving of alcohol
⃞ calculate the BAC for males and females
⃞ calculate the time it takes for a person’s BAC to fall to zero
⃞ identify and explain the limitations of methods used to estimate BAC
⃞ calculate medical dosages using Fried’s, Young's and Clark’s formulas
⃞ change the subject of a formula that is a simple quadratic equation
⃞ assess whether a solution is reasonable in the context of the problem
⃞ use a spreadsheet to perform calculations involving formulas
Related resources
Review quiz: Formulas and equations
I need to review this
Go back to Lesson 1.1 Substitution into expressions and formulas
Go back to Lesson 1.2 Solving linear equations
Go back to Lesson 1.3 Solving further linear equations
Go back to Lesson 1.4 Variables and formulas
Go back to Lesson 1.5 Travel calculations
Go back to Lesson 1.6 Blood alcohol content calculations
Go back to Lesson 1.7 Medication calculations
Go back to Lesson 1.8 Solving simple quadratic equations
